Q

# Help me please, - Current Electricity - JEE Main

A metal wire of resistance  $3\:\Omega$ is elongated to make a uniform wire of double its previous length . This new wire is now bent and the ends joined to make a circle . If two points on this circle make an angle at the centre , the equivalent resistance between these two points will be :

• Option 1)

$\frac{12}{5}\Omega$

• Option 2)

$\frac{5}{12}\Omega$

• Option 3)

$\frac{5}{3}\Omega$

• Option 4)

$\frac{7}{2}\Omega$

Views

$R_{initial }=3\: \Omega$

After elongation

Vol of wire initial = vol of wire  final

$\Rightarrow\pi\: r_{i}^{2}l_{i}=\pi r_{j}^{2}\:2\:l_{i}$

$\Rightarrow \:\:r_{j}^{2}=\frac{r_{i}^{2}}{2}$

$R_{i}=\rho \:\frac{l}{A}=\rho \:\frac{l_{i}}{\pi\:r_{i}^{2}}$

$R_{final}(after \:\:elongation)=\rho \:\frac{2l_{i}}{\pi\frac{r_{i}^{2}}{2}}=4Ri=12 \Omega$

$R_{1}= \frac{\left ( \frac{5\pi}{3} \right )}{2\pi}\times 12= 10 \Omega$

$R_{2}= \frac{\left ( \frac{\pi}{3} \right )}{2\pi}\times 12= 2 \Omega$

$R_{eq}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}=\frac{5}{3} \Omega$

Option 1)

$\frac{12}{5}\Omega$

Option 2)

$\frac{5}{12}\Omega$

Option 3)

$\frac{5}{3}\Omega$

Option 4)

$\frac{7}{2}\Omega$

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